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Thread: Why +2lbs On The Cross?

04102005, 11:55 AM #35
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Originally Posted by ants
Did it play differently?

04212005, 06:57 AM #36
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a question i've thought of
If stringing at 2 lbs difference , won't the course of the shuttle differ when the shuttle hit the area where there's crosses and mains due to the fact that the amount that the string would be able to stretch differs from the different tension ?

04212005, 06:06 PM #37
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Originally Posted by tangle
hope this helps

04212005, 11:21 PM #38
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27 lbs X 27 lbs on a MP100
Last night I bumped into an unregistered BF lurker at VRC. He showed me his MP100 that was outwardly round cross wise. It was initially strung at 27 lbs X 27 lbs and its shape was all right. I guess after a couple of hard sessions the cross strings relaxed more than the main strings. So, a safe bet is still to add 2 extra pounds on the cross to keep shape of the racquet. Just my two cents.

04222005, 02:20 AM #39
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Originally Posted by Pete LSD
Maybe it was done on a 2 point machine.
My machine has side supports.
I have been stringing my own racquets with the cross at lower tension than the mains since Jan 2005 and have had no problems.
Mostly at 22x20, but i have had my Ti10 and MP100 at 26x24 and 26x25 and not seen any problems with the head shape.

04222005, 01:47 PM #40
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This dude's stringer uses a crank machine. Not sure if it has side supports.
Originally Posted by Neil Nicholls

04302005, 06:37 AM #41
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Originally Posted by ants
What did you find?
Did it play differently?

05262005, 07:22 AM #42
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reduce the main string 2 lbs
when a customer wants restring 23lbs, it's mean main 21lbs and cross 23lbs, but not all cross string is 23lbs, it's from head's 6th staring string, it's mean from the sweatspot to bottom.

06112005, 09:48 AM #43
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String tension
Hello everybody,
I'm stringing my racket also with 510% less tension on the crosses.
Somewhere I was reading an Article in the net that the recommendation ( stringing the crosses 2 lbs higher) is only because the deformation of the headshape.
The main aim should be gaining an equal stringbed tension. Therefore because the crosses are shorter than the mains they have to be strung with a lower tension.
I tried to proof this with some physics. Here are my calculations
Because the string lenghten itself while under tension the diameter => the cross sectional area of the string is getting smaller.
A = V /(L1 * k * F1) ; A = Cross sectional Area, V = Volume of the unstretched String,
k = lenghtening of the string (%), F1 Force (N) which pulls on the string
the tension of the string is coupled with it's lenght through:
2 * L1 * f = sqrt(F1/ A * ro) ; f = Frequency, ro = density (constant)
(c = sqrt(F1/ A * ro) ; , c = propagationspeed of the wave
with c= lambda f; lambda = wave length = 2 *L1)
The frequency ( Sound) of the long (main L2) and the cross (L1) string should be the same, therefore we can insert the two equations into each other and get :
L1/L2 = sqrt[(F1* F1 * L1 * k * v2 * ro)/ (F2 * F2 * L2 * k * v1 * ro)]= sqrt([v2 * L1]/v1 * L2]) * F1/F2
we can cancel some terms and get:
=> F1 = F2 * (sqrt(v2 * L2/ v1* L1))^1 ; with v1 = (d/2)^2 * pi * L1 , d diameter of
the string
the equation simplifies to:
F1 = F2 * ( L1/ L2)
If we take L1 = 18,8 cm for the cross string and L2 = 23,5 cm for the main strings
we get a tension ratio of:
F1/F2 = L1/L2 = 0.8
If the mains are strung with 22lbs (10kg) the crosses have to be strung with 17.6 lbs (8kg) to get the same overall stringbedtension.
Of course this is just for the longest strings, but I think it should be clear now that the crosses should have a lower tension.
Hopefully there are no mistakes in my equations but the result is just to nice to be wrong.
Lets hear what you think about it.
Regards mark

06142005, 04:11 AM #44
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Why 2 more lbs on the crosses
Hello,
it seems as nobody is reeding my post. So I had to find an error myself. The results were just to nice
I think I have an error in the equation to calcualte the cross sectional area A in dependence of the Force, because if you apply zero force the length of the string would be zero!
A = V /(L1 * k * F1) ; A = Cross sectional Area, V = Volume of the unstretched String,
has to be corrected to:
A = V /(L1 * (1+ k * F1))
the rest follows analogue to my previous post. But the result is not as nice as before. The equation for the wanted string tension (Force) is:
F2 = 1/2K2 * [sqrt(1+4k2(L2R2/L1R1)^2*F1*(1+kF1))1)
Hopefully the equation is now correct
If we are using the relative lenghtening from Neils post for BG 65 (0,70cm) 17% and 23.5cm for the mains, 18.5cm for the crosses and the applied force is 100 N ( roughly 10kg = 22lb), the the tension for the crosses should be 78,1N or 7,8kg which is nearly the same ratio as before (0,8)
With the computed equation it's also possible to calculate the necesseary tension for the proportinal stringing method or the resulting tension if you are using an other string.
For example if you are using normally the BG65 and string this one with 22lb and use this tension for BG68TI the equivalent tension with BG65 would be 31 lb!! Due to the extreme lenghtening of the string 33 %.
If there are still some errors I would be glad to get some corrections.
Regards mark

06142005, 05:33 AM #45
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Originally Posted by baumbaer
so for 10kg tension, the string would stretch 11.7%

06152005, 01:38 AM #46
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why 2 lbs more on the cross
Originally Posted by Neil Nicholls
Perhaps someone is able to verify the results.
Regards mark

06152005, 01:52 AM #47
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hi mark
all u have done was deriving the basic ratio quotient the long way
x1/x2=y1/y2
actually, equal main and cross tension is just one of many objective desired by the stringer/userLast edited by cooler; 06152005 at 02:03 AM.

06152005, 02:21 AM #48
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Originally Posted by Neil Nicholls

07012005, 03:32 PM #49
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i just popped into this forum section after reading the equipment section...
and i noticed this thread... adding 2 pounds to the crosses to maintain the stringbed's normal shape so it isnt deformed. therefore, i'm going to string my new mp99 with bg66@main 22lbs and cross 24lbs? but what will the tension be? both equal since the mains are longer? therefore both will be at 22pounds, or what?
and also the mathmatical equation by baumbaer and neil.
to put it in short,the conclusion you have reached is that if one kind of string is done at a certain tension, another string will have to be at a different tension to be exactly the same tension after stretching? correct?
very complicated indeed...and im making no sense of that mathmatical equation...

07022005, 04:05 AM #50
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Originally Posted by cooler

07022005, 11:39 AM #51
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i want my racquet to be strung at 24 lbs even
therefore, with this pattern, i would string it at
24@ mains
26@ crosses
after a couple ours of rest, then it would relax to 24 together?
btw, is the mp99 string pattern 22 mains 22 crosses?
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